calculus derivative problem help

Hi, I need help finding the derivative for the problem h(x)= x/2^(1/2) + cos x - 2^(1/2)/2. I think the derivative of cos x is -sin x and -2^(1/2)/2 should be 0 because of the constant rule but what are the steps for the derivative of the first part of the problem? The book I'm using only gives the answer and no explanation. Thank you.

Re: calculus derivative problem help

We are given:

$\displaystyle h(x)=\frac{1}{2^{\frac{1}{2}}}x+\cos(x)-\frac{2^{\frac{1}{2}}}{2}$

Differentiate term by term. For the first term use:

$\displaystyle \frac{d}{dx}(kx)=k$

For the second term:

$\displaystyle \frac{d}{dx}(\cos(x))=-\sin(x)$

And for the third term:

$\displaystyle \frac{d}{dx}(k)=0$.

In the above formulas, *k* is an arbitrary constant.

Re: calculus derivative problem help

If the function is exactly what you have written then it is the form h(x)= ax+ cos(x)- a with a a constant: $\displaystyle a= \sqrt{2}/2$. Do you know how to differentiate ax+ cos(x)- a?